Consider point a which is 70 cm north of a -3.4 µC point charge, and point b which is 82 cm west of the charge?

a)Determine Vba = Vb - Va. Answer in Volts

b)Determine MAGNITUDE of Eb - Ea. Answer in N/C

c)Direction of Eb-Ea. Answer in Degrees. (counterclockwise from east is positive)

1 Answer

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  • Ossi G
    Lv 7
    9 years ago
    Favorite Answer

    a)

    potential at A = kQ/r = 9*10^9*(-3.4*10^-6)/0.7 = - 43714 V

    potential at B = 9*10^9(-3.4*10^-6)/0.82 = - 37317 V

    Vba = 6397 V

    b)

    Ea = 9*10^9*3.4*10^-6/0.7^2 = 62449 N/C pointing south

    Eb = 9*10^9*3.4*10^-6/0.82^2 = 45508 N/C pointing east

    Eb - Ea = (45508, - 62449)

    magnitude = √(62449^2+45508^2)= 77271 N/C

    pointing tan^-1(62449/45508) = +53.92°

    O:G:

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